Proving that $n!≤((n+1)/2)^n$ by induction I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this: 
$V(1): 1≤1 \text{  true}$ 
$V(n): n!≤((n+1)/2)^n$ 
$V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}$
and I've got : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$
 A: It is more easy to prove this inequality without induction. Really $$0 < i\cdot (n + 1 - i) = \left(\frac{n+1}2 + \frac{2i - n - 1}2\right)\left(\frac{n+1}2 - \frac{2i - n - 1}2\right) = \left(\frac{n+1}2\right)^2 - \left(\frac{2i - n - 1}2\right)^2 \le \left(\frac{n+1}2\right)^2.$$
Multiply this inequalities for all $i = 1, 2, \ldots, \left\lfloor\frac n2\right\rfloor$ and by $\frac{n+1}2 = \frac{n+1}2$ for odd $n$ to get $n! \le \left(\frac{n+1}2\right)^n$ as desired.
A: An induction proof:
First, let's make it a little bit more eye-candy:
$$
n! \cdot 2^{n} \leq (n+1)^n
$$
Now, for $n=1$ the inequality holds. For $n=k\in\mathbb{N}$ we know that:
$$
k! \cdot 2^{k} \leq (k+1)^k
$$
holds and we need to prove:
$$
(k+1)! \cdot 2^{k+1} \leq (k+2)^{k+1}
$$
We will now prove this chain of inequalities (which gives us the actual proof):
$$
(k+1)! \cdot 2^{k+1} \leq 2(k+1)^{k+1} \leq (k+2)^{k+1}
$$
The first inequality is from the assumption (both sides multiplied by $2(k+1)$). Now we just need to prove the second one. In other words, we need to prove this (for some big enough positive integer $p$):
$$
2p^{p} \leq (p+1)^{p}
$$
And that's rather obvious. The inequality
$$
2 \leq \left(1+\frac{1}{p}\right)^{p}
$$
holds because the function on the right is known to be increasing and its limit (as $p\to\infty$) is $e$. So at some point on it has to be greater than 2.
A: If you really need induction let it be.
Base is $n = 0$: $0! = 1 \le 1 = \left(\frac12\right)^0$.
By induction hypothesis the inequality holds for $n = k$. Let proove it for $n = k+1$.
$$k! \le \left(\frac{k+1}2\right)^k,\\
k!(k+1) \le \left(\frac{k+1}2\right)^k(k+1),$$
$$(k+1)! \le \frac{(k+1)^{k+1}}{2^k}.\tag{*}$$
Now we need to show that $f(x) = \frac{x^{x}}{(x+1)^x} \le \frac12$. Ok, $f(x) = \left(\frac{x}{x+1}\right)^x = e^{x(\ln x - \ln (x+1))}$, then
$$f'(x) = e^{x(\ln x - \ln (x+1))}\left((\ln x - \ln (x+1)) + x\left(\frac1x - \frac1{x+1}\right)\right) =
e^{x(\ln x - \ln (x+1))}\left(\ln \left(1 - \frac1{x+1}\right) + \frac1{x+1}\right) \le 0$$ for any positive $x$, since $\ln y \le y - 1$ for any positive $y$. So $f(x) \le f(1) = 1/2$ for any $x \ge 1$. Then from (*) we get
$$(k+1)! \le \frac{(k+1)^{k+1}}{2^k} \le \frac{(k+1)^{k+1}}{2^k}\cdot \frac1{2f(k+1)} = \frac{(k+1)^{k+1}}{2^k}\cdot \frac{(k+2)^{k+1}}{2(k+1)^{k+1}} = \frac{(k+2)^{k+1}}{2^{k+1}} = \left(\frac{k+2}2\right)^{k+1}.$$
